hdu 4617 2013多校联合训练第二场weapon简单的计算几何

#include<cstdio>
#include<algorithm>
#include<cmath>
#define INF 1e9
#define eps 1e-8
using namespace std;
struct Point {double x,y,z;};
typedef Point Vec;
struct Yuanzhu{double r;Vec N;Point center;}team[50];
double inline Veclen(Vec u)//计算模长
{
return  sqrt(u.x*u.x+u.y*u.y+u.z*u.z);
}
Point inline Jian(Point a,Point b)//2个点相减
{
return (Point){a.x-b.x,a.y-b.y,a.z-b.z};
}
double inline DianJi(Vec a,Vec b)//点积
{
return a.x*b.x+a.y*b.y+a.z*b.z;
}
Vec inline ChaJi(Vec a,Vec b)//叉积
{

return (Vec){a.y*b.z-b.y*a.z,b.x*a.z-a.x*b.z,a.x*b.y-b.x*a.y};
}
bool inline Zero(double a)//浮点判零
{
return a<eps&&a>-eps;
}
double dis(Yuanzhu a,Yuanzhu b)//算2个圆柱间的距离
{
Vec AB=Jian(a.center,b.center);//圆心连接的构成的向量
Vec N=ChaJi(a.N,b.N);
if(Zero(N.x)&&Zero(N.y)&&Zero(N.z)) N=a.N;//如果2个圆柱平行
double ans;
ans=DianJi(AB,N)/Veclen(N);//计算2个圆柱的过圆心的直线的最小距离，注意可能为负值
if(ans<0) ans=-ans;//为负值就取其绝对值
return ans-a.r-b.r;//减去2个圆柱的半径，就得到了2个圆柱的最小距离，为负值的时候说明2个圆柱相交，0的时候相切
}
void inline scan(Point &a) { scanf("%lf%lf%lf",&a.x,&a.y,&a.z); }//读入一个点的函数
int n;
void read()
{
scanf("%d",&n);
for(int i=0;i<n;i++)
{
Point a,b;
scan(team[i].center);scan(a);scan(b);
team[i].r=Veclen(Jian(team[i].center,a));
team[i].N=ChaJi(Jian(a,b),Jian(team[i].center,a));
}
}
void deal()
{
double ans=INF;
for(int i=0;i<n;i++)
for(int j=i+1;j<n;j++)//枚举2个圆柱的距离
{
double temp=dis(team[i],team[j]);
if(temp<eps)//判断是否相交
{
printf("Lucky\n");
return ;
}
ans=min(ans,temp);
}
printf("%.2lf\n",ans);
}
int main()
{
int T;scanf("%d",&T);
while(T--)
{
read();
deal();
}
return 0;
}